A body covers 200 c m in the first 2 seconds and 220 c m in the next 4 seconds. The velocity of the body at the end of the 7 t h second is :

Question

a. 40m/s

b. 200m/s

c. 10m/s

d. 5m/s

Answer 1

The question is unanswerable unless we assume constant acceleration.

1st part: d₁ = v₀t₁ + (1/2) at₁² or .20 = 2v₀ + 2a or .10 = v₀ + a

also you know v₁ = v₀ + at = v₀ + 2a at t=2

2nd part d₂ = v₁t₂ + (1/2)at₂² = (v₀+2a)(4) + (1/2)a(16) = 4v₀ + 16a = .22m

Subtract 4 x d₁ equation from d₂ equation to solve for a: (16-4)a = .22-.4 a = -.015 m/s² and v₀ = .115 m/s

For t=7s, just plug t =7 into the d₁ equation.

You could have gotten a cleverly ig you know that the average velocity in a constant acceleration situation applies at a time between the two times: v_avg,1 = .2/2 = .1 at t = 1 and v_avg,2 = .22/4 = .055 at t = 4 from (2+6)/2, not the 4 second interval. So the a =( v_avg,1 - v_avg,2)/(4-1) = -.015 m/s², then v₀ , then v(7) can be solved for.

Please consider a tutor. take care.